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Resistors in seriess-current is the same?

  1. Mar 20, 2014 #1
    resistors in seriess---current is the same??

    So current is the same if you have resistors in series….but won't the current decrease because the resistors provide resistance to the current??\
  2. jcsd
  3. Mar 20, 2014 #2


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    Isn't current inversely proportional to resistance for a set voltage? I mean, adding up the resistance of resistors in series is just R(total) = r1 + r2 + r3 but that doesn't change anything about Ohm's law. You would have to add more voltage to get the same current flow. (I would think)
  4. Mar 20, 2014 #3
    Question is not clear but I interpret it as being about why when two resistors - say R1 and R2 - are in series the current through R1 is the same as the current through R2. If they were different than some of the electrons passing through R1 would not pass through R2. Where do you suppose they would go? If I were to connect two straws in series than it might get harder for me to pump water through them - increased resistance - and the flow of water - the current - would decrease. But the flow through the first straw would be the same as the flow through the second straw. Otherwise where do we suppose the water would go? If I increase the pressure difference between the ends of the straw - increased voltage - the flow of water through the straw increases - increased current. That's the essence of Ohm's law.
  5. Mar 20, 2014 #4
    But isn't current coulombs per second?? So it measures the rate at which the current is flowing? So wouldn't the resistor "hold back" some charge, making it slow down --> less current
  6. Mar 20, 2014 #5
    No, the resistor doesn't hold back any charge. All the charge that goes in at one end comes out at the other end. It just takes longer to get there. Just as all the water that goes in at one end of a straw comes out at the other end. If you squeeze the straw - increasing the resistance - all the water that goes in still comes out at the other end, but it takes longer to get there.
  7. Mar 20, 2014 #6


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    That isn't a very good analogy because, as you said, it takes the water more time to go through the straw, making the current q/s, smaller. Even so, the resistor doesn't "slow down" the current when the circuit is in equilibrium.

    Edit: +1 on post #3
  8. Mar 20, 2014 #7
    It is an excellent analogy. You must've misunderstood post #3. The resistor does slow down the motion of the electrons reducing the current just as a constriction in a straw slows down the flow of water.
  9. Mar 21, 2014 #8


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    The current flowing into a resistor is the same as the current flowing out - only the voltage changes, which is the essence of Ohm's Law: voltage drop = current x resistance.

    So if you put two resistors in series the current remains the same.

    You can understand this in terms of "where does the charge go?" - capacitors store charge, but resistors (ideal resistors) don't ... hence every bit of charge entering must exit.

    Remember, a wire is also a resistor ... they may have a low resistance, but it is cumulative over the length of a wire.
  10. Mar 21, 2014 #9
    what if i tell you it doesn't ? well usually it does but not necessarily , actually in circuits with different resistors , current flows faster through resistors of low Cross section area , to maintain a steady current , anyway its NOT about speed of electrons
    back to OP
    the way i like to think about it is as follows
    imagine a train that is composed of 3 carriages , as the train was moving , you exert a force in the carriage in the middle that slows it down * lets for the sake of our argument consider the velocity of the train to be analogous to the current intensity * anyway , as you slow down the carriage in the middle , what is going to happen ?
    since the carriages are connected with some bond that can neither expand nor compress
    once you slow down the carriage in the middle the carriage will slow down the other carriages , the force will be propagated quickly to the carriage in frfont of it and the carriage behind it , slowing them down just as much as the carriage in the middle was slowed down , thus the train carriages keep moving with the same velocity
    now back to the current point of view
    forgive me for connecting ideas of current and velocity but it was just to serve the purpose of the analogy , anyway
    in the current , suppose you have 3 resistors , then a fourth resistor appear out of no where , what is going to happen is that at the point where the charges enter this 4th resistor , the current will decrease at this point where it is forced to enter a resistor , thus causing a build up of charge that happens in an infinitesimal amount of time , this build up of charges pushes the charges behind it , and these charges push the charges behind them and it goes on to the start point of the circuit .
    of course all of this happens in very small amount of time that is too small to be noticed .
    just like the 3 carriages , if you slow down the first carriage , it will cause the coupling between the first carriage and the second one to compress , and of course since the coupling is sturdy it will not be compressed * or it might compress for a very small amount of time * so the force propagates through the coupling and slow down the second carriage , and the same happens to the third carriage .
    hope this helped

    btw in this analogy , the coupling is analogous to the electric force between electrons
    and in case you didnt know what a coupling is
  11. Mar 21, 2014 #10


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    It sounds like you're confusing two different situations. In a given circuit, if two resistors are in series, the current through the first resistor is the same as the current through the second resistor. You, however, are thinking of two different circuits, one with only one resistor and a second circuit in which you add a resistor in series with the first resistor. In this case, yes, the current decreases because the total resistance increases.
  12. Mar 21, 2014 #11


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    Analogies 'mon cerveau fait mal'.
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